\section[$TE_z$ Radial Modes]{Transverse Electric $TE_z$ Radial Modes}
In order to obtain $TE_z$ modes, the following conditions of the vector potentials are
\begin{align}
\vec{F} &= F_z\hat{z}\\
\vec{A} &= 0
\end{align}
Using separation of variables, the electric vector potential is defined as
\begin{align}
F_z(\rho,\phi,z) &= {C_f}\left[{\sqrt{Z_{f}}}v_F(\rho)\right]\Phi_F(\phi)Z_F(z)\label{eqn:Fzcyl}\\
&= {C_f}[V(\rho)]\Phi_F(\phi)Z_F(z)\label{eqn:Fzcyl2}
\end{align}
where a $TE_\rho$ impedance is defined as,
\begin{align}
Z_{f} = \frac{j\omega\mu'}{k_{f\rho}}\label{eqn:ZTErho}
\end{align}
and where $C_f$ is a normalization constant.  The solutions must satisfy the wave equation
\begin{align}
(\nabla^2+k^2)F_z=0
\end{align}
where
\begin{align}
k^2=k_{f\rho}^2+k_{fz}^2\label{eqn:constrainteqncyl}
\end{align}
which is the \emph{constraint equation} in cylindrical coordinates.

Since there are perfect electric conductors (pec) on the top and bottom of the cylinder, $Z_F(z)$ will be expanded as trigonometric functions which will form standing waves between the top and bottom plates.  The function and its derivative are defined as,
\begin{align}
Z_F(z) &= C_{fe}\cos(k_{fz}z)+C_{fo}\sin(k_{fz}z)\label{eqn:Zcyl}\\
Z'_F(z) &= \frac{\partial{Z_F(z)}}{\partial{z}}= -k_{fz}\left(C_{fe}\sin(k_{fz}z)-C_{fo}\cos(k_{fz}z)\right)\label{eqn:dZcyl}
\end{align}

The fields in the $\hat{\phi}$ direction also form standing waves so they will be expanded as trigonometric functions in which it and its derivative are given as, 
\begin{align}
\Phi_F(\phi) &= D_{f0}\cos(\nu\phi)+D_{f1}\sin(\nu\phi)\nonumber\\
&=\sum_{\ell=0}^{1}D_{f\ell}\cos\left(\nu\phi-\ell\frac{\pi}{2}\right)\label{eqn:Phicyl}\\
\Phi'_F(\phi) &= \frac{\partial{\Phi_F(\phi)}}{\partial{\phi}}= -\nu\left(D_{f0}\sin(\nu\phi)-D_{f1}\cos(\nu\phi)\right)\nonumber\\
&= -\nu \sum_{\ell=0}^{1}D_{f\ell}\sin\left(\nu\phi-\ell\frac{\pi}{2}\right)\label{eqn:dPhicyl}
\end{align}
where in general $\nu$ is a real number.

Since the fields are assumed to be propagating in the $\hat{\rho}$ direction, $R_F(\rho)$ will be expanded as forward and backward traveling waves with constant coefficients, which are most conveniently described as \emph{Hankel} functions in the cylindrical coordinate system.  The function along with it's derivative are given as,
\begin{align}
v_F(\rho) &= R_F^+H_\nu^{(2)}(k_{f\rho}\rho)+R_F^-H_\nu^{(1)}(k_{f\rho}\rho)\label{eqn:Rcyl}\\
R_F'(\rho)&=\frac{d{R_F(\rho)}}{d{\rho}}=R_F^+{H_{\nu}^{(2)}}'(k_{f\rho}\rho)+R_F^-{H_{\nu}^{(1)}}'(k_{f\rho}\rho)\label{eqn:dRcyl}
\end{align}
where 
\begin{align}
{H_{\nu}^{(2)}}'(k_{f\rho}\rho)&=\frac{d}{d\rho}\left[{H_{\nu}^{(2)}}(k_{f\rho}\rho)\right]=-k_{f\rho}H_{{\nu}+1}^{(2)}(k_{f\rho}\rho)+\frac{\nu}{\rho}H_{\nu}^{(2)}(k_{f\rho}\rho)\label{eqn:Hnu2}\\
{H_{\nu}^{(1)}}'(k_{f\rho}\rho)&=\frac{d}{d\rho}\left[{H_{\nu}^{(1)}}(k_{f\rho}\rho)\right]=-k_{f\rho}H_{{\nu}+1}^{(1)}(k_{f\rho}\rho)+\frac{\nu}{\rho}H_{\nu}^{(1)}(k_{f\rho}\rho)\label{eqn:Hnu1}
\end{align}
The variable $\nu$ is the order of the Hankel function and is the variable that couples $R_F(\rho)$ and $\Phi_F(\phi)$.

\subsection{$TE_z$ Field Solutions}
For $TE_z$ modes, the electric and magnetic fields are found by substituting (\ref{eqn:Fzcyl}) into (\ref{eqn:EnormTE})--(\ref{eqn:HperpTE}) and expanding in the the cylindrical coordinate system to get,
\begin{align}
\vec{E}_F(\rho,\phi,z) &= C_f\sqrt{Z_f}\left[-\frac{1}{\rho}R_F(\rho)\Phi'_F(\phi)Z_F(z)\hat{\rho}+R'_F(\rho)\Phi_F(\phi)Z_F(z)\hat{\phi}\right]\label{eqn:ETEcyl}
\end{align}
\begin{multline}\label{eqn:HTEcyl}
\vec{H}_F(\rho,\phi,z) = \frac{C_f}{\sqrt{Z_f}}\frac{1}{k_{f\rho}}\biggl[R'_F(\rho)\Phi_F(\phi)Z'_F(z)\hat{\rho}\\+\frac{1}{\rho}R_F(\rho)\Phi'_F(\phi)Z'_F(z)\hat{\phi}+k_{f\rho}^2R_F(\rho)\Phi_F(\phi)Z_F(z)\hat{z}\biggr]
\end{multline}
where $'=\frac{\partial}{\partial{u}}$ with $u=\{\rho,\phi,z\}$. Solving for $k_{f\rho}$ in (\ref{eqn:constrainteqncyl}) yields,
\begin{align}
k_{f\rho} = -j\sqrt{k_{fz}^2-k^2}\label{eqn:krho}
\end{align}
in which the negative sign was chosen deliberately on the square root to ensure waves whose magnitude approach zero in the direction of propagation.
\subsection{Boundary Conditions}
The field variations in the $\phi$ direction are periodic such that,
\begin{align}
\Phi_F(\phi)=\Phi_F(\phi+2\pi)
\end{align}
Since the fields are periodic then the fields must repeat every $2\pi$ so that $\nu=m$ where,
\begin{align}
m= \left\{ \begin{array}{ll}
0,1,2,\ldots & \textrm{if $\ell=0$}\\
1,2,3,\ldots & \textrm{if $\ell=1$}\\
\end{array} \right.
\end{align}
which can also be defined with the intermediate variable $m'$ as
\begin{align}
m'&= 0,1,2,\ldots\\
m &= m'+\ell
\end{align}

The boundary conditions in the $\hat{z}$ direction will now be applied to solve for $k_{fz}$ and either $C_{fe}$ or $C_{fo}$. The first boundary condition applied will be to the bottom plate of the cylinder where
\begin{align}
E^F_{\rho}(z=0)=E^F_{\phi}(z=0)=0
\end{align}
which from (\ref{eqn:ETEcyl}) implies that 
\begin{align}
Z_F(0) &= C_{fe}\cos(k_{fz}0)+C_{fo}\sin(k_{fz}0)=C_{fe}=0
\end{align}
Application of the boundary condition on the top plate
\begin{align}
E^F_{\rho}(z=z_0)=E^F_{\phi}(z=z_0)=0
\end{align}
implies that 
\begin{align}
Z_F(z_0) &= C_{fo}\sin(k_{fz}z_0)=0
\end{align}
and since $k_{fz}\ne0$ then $k_{fz}z_0={n}\pi$, or
\begin{align}
k_{fzn} = \frac{{n}\;\pi}{z_{0}}
\end{align}
where ${n}=1,2,\ldots\infty$.  Equation (\ref{eqn:Zcyl}) and (\ref{eqn:dZcyl}) now reduce to,
\begin{align}
Z_{F}(z)_n &= \sin\left(k_{fzn}z\right)\label{eqn:Zcyl2}\\
Z'_{F}(z)_n &= k_{fzn}\cos\left(k_{fzn}z\right)\label{eqn:dZcyl2}
\end{align}
Also the separated functions (\ref{eqn:Phicyl}) and (\ref{eqn:dPhicyl}) change to,
\begin{align}
\Phi_F(\phi)_{\ell,m} &=\cos\left(m\phi-\ell\frac{\pi}{2}\right)\label{eqn:Phicyl2}\\
\Phi'_F(\phi)_{\ell,m}&=-m\sin\left(m\phi-\ell\frac{\pi}{2}\right)\label{eqn:dPhicyl2}
\end{align}
in which $C_{fon}=D_{f\ell,m}=1$.
Equations (\ref{eqn:Rcyl}) and (\ref{eqn:dRcyl}) become,
\begin{align}
R_F(\rho)_{\ell,m,n} &= R_{F\ell,m,n}^+H_{m}^{(2)}(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-H_{m}^{(1)}(k_{f\rho{n}}\rho)\label{eqn:Rcyl2}\\
R'_F(\rho)_{\ell,m,n} &= R_{F\ell,m,n}^+{H_{m}^{(2)}}'(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{m}^{(1)}}'(k_{f\rho{n}}\rho)\label{eqn:dRcyl2}
\end{align}
where 
\begin{align}
{H_{m}^{(2)}}'(k_{f\rho{n}}\rho)&=\frac{d}{d\rho}[{H_{m}^{(2)}}(k_{f\rho{n}}\rho)]=-k_{f\rho n}H_{m+1}^{(2)}(k_{f\rho n}\rho)+\frac{m}{\rho}H_{m}^{(2)}(k_{f\rho n}\rho)\label{eqn:dHm2cyl2}\\
{H_{m}^{(1)}}'(k_{f\rho{n}}\rho)&=\frac{d}{d\rho}[{H_{m}^{(1)}}(k_{f\rho{n}}\rho)]=-k_{f\rho n}H_{m+1}^{(1)}(k_{f\rho n}\rho)+\frac{m}{\rho}H_{m}^{(1)}(k_{f\rho n}\rho)\label{eqn:dHm1cyl2}
\end{align}

\subsection{Basis Set and Field Expansion}
After applying boundary conditions, the solutions are seen to be a discrete set of modes, and because of linearity, the total electric and magnetic fields are found by summing (i.e., superposition) each of the individual modes to get
\begin{multline}\label{eqn:EFcyl}
\vec{E}_F(\rho,\phi,z)=\sum_{{\ell}=0}^1\sum_{{m}=0}^\infty\sum_{{n}=1}^\infty C_{f\ell,m,n}\sqrt{Z_{fn}}\biggl[-\frac{1}{\rho}R_F(\rho)_{\ell,m,n}\Phi'_F(\phi)_{\ell,m}Z_F(z)_n\hat{\rho}\\+R'_F(\rho)_{\ell,m,n}\Phi_F(\phi)_{\ell,m}Z_F(z)_n\hat{\phi}\biggr]
\end{multline}
\begin{multline}\label{eqn:HFcyl}
\vec{H_F}(\rho,\phi,z)=\sum_{{\ell}=0}^1 \sum_{{m}=0}^\infty \sum_{{n}=1}^\infty \frac{C_{f\ell,m,n}}{\sqrt{Z_{fn}}}\frac{1}{k_{f\rho{n}}}\biggl[R'_F(\rho)_{\ell,m,n}\Phi_F(\phi)_{\ell,m}Z'_F(z)_n\hat{\rho}\\
+\frac{1}{\rho}R_F(\rho)_{\ell,m,n}\Phi'_F(\phi)_{\ell,m}Z'_F(z)_n\hat{\phi}+k_{f\rho{n}}^2R_F(\rho)_{\ell,m,n}\Phi_F(\phi)_{\ell,m}Z_F(z)_n\hat{z}\biggr]
\end{multline}
Each function and variable was tagged with an index, where each individual mode will have a triple of indices $\ell,m,n$. Therefore, a mode is distinguished from all other modes by this index triple. From (\ref{eqn:EFcyl}) and (\ref{eqn:HFcyl}) a basis set is defined, where the $\hat{\rho}$ components are given by
\begin{align}
e^F_{\rho}(\rho,\phi,z)_{\ell,m,n}&=\frac{C_{f\ell,m,n}}{\rho}\sqrt{Z_{fn}}\Phi'_F(\phi)_{\ell,m}Z_F(z)_n\\
h^F_{\rho}(\phi,z)_{\ell,m,n}&=\frac{-C_{f\ell,m,n}}{\sqrt{Z_{fn}}}\Phi_F(\phi)_{\ell,m}Z'_F(z)_{n}
\end{align}
and where the tangential components are defined as
\begin{align}
\vec{e}_F(\phi,z)_{\ell,m,n} &= -C_{f\ell,m,n}\sqrt{Z_{fn}}k_{f\rho n}\Phi_F(\phi)_{\ell,m}Z_F(z)_{n}\hat{\phi}\label{eqn:EFbasiscyl}\\
\vec{h}_F(\rho,\phi,z)_{\ell,m,n} &= \frac{C_{f\ell,m,n}}{\sqrt{Z_{fn}}}
\left[\frac{1}{k_{f\rho{n}}\rho}\Phi'_F(\phi)_{\ell,m}Z'_F(z)_{n}\hat{\phi}+k_{f\rho{n}}\Phi_F(\phi)_{\ell,m}Z_F(z)_{n}\hat{z}\right]\label{eqn:HFbasiscyl}
\end{align}

The total $TE_z$ field components normal to the $\hat{\rho}$ direction, expanded in terms of the basis set are given as,
\begin{align}
E^F_{\rho}(\rho,\phi,z)&=\sum_{\ell,m,n}{e}^F_{\rho}(\phi,z)_{\ell,m,n}\left[R_{F\ell,m,n}^+{H_{m}^{(2)}}(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{m}^{(1)}}(k_{f\rho{n}}\rho)\right]\\
H^F_{\rho}(\rho,\phi,z)&=\sum_{\ell,m,n}{h}^F_{\rho}(\phi,z)_{\ell,m,n}\frac{-1}{k_{f\rho{n}}}\left[R_{F\ell,m,n}^+{H_{m}^{(2)}}'(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{m}^{(1)}}'(k_{f\rho{n}}\rho)\right]
\end{align}
where the tangential fields expanded in terms of the basis set are given as,
\begin{align}
\vec{E}^F_{\bot}(\rho,\phi,z) &= \sum_{\ell,m,n}\vec{e}_F(\phi,z)_{\ell,m,n}\frac{-1}{k_{f\rho n}}\left[R_{F\ell,m,n}^+{H_{m}^{(2)}}'(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{m}^{(1)}}'(k_{f\rho{n}}\rho)\right]\label{eqn:EFperpcyl}\\
\vec{H}^F_{\bot}(\rho,\phi,z) &= \sum_{\ell,m,n}\vec{h}_F(\rho,\phi,z)_{\ell,m,n}\left[R_{F\ell,m,n}^+{H_{m}^{(2)}}(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{m}^{(1)}}(k_{f\rho{n}}\rho)\right]\label{eqn:HFperpcyl}
\end{align}
It should be noted that since the tangential magnetic fields are dependent on the variable $\rho$, or in other words $\rho$ cannot be factored out, then these basis fields are unfortunately not suitable for a scattering matrix formulation.

\subsection{$TE_z$ Radial Symmetry (Horizontal Polarization)}
If the structure has radial symmetry (no variation in the $\phi$ direction) then
\begin{align}
\Phi'_F(\phi)=0
\end{align}
which makes 
\begin{align}
\Phi_F(\phi) = c
\end{align}
where $c$ is a constant which will be set to unity.  Under these conditions the index $m$ and $\ell$ reduce to $m=\ell=0$.  
The other two separated variables reduce to,
\begin{align}
Z_F(z)_n &= \sin\left(k_{fzn}z\right)\\
Z'_F(z)_n &= k_{fzn}\cos\left(k_{fzn}z\right)
\end{align} 
\begin{align}
R_F(\rho)_n &= R_{Fn}^+H_0^{(2)}(k_{f\rho{n}}\rho)+R_{Fn}^-H_0^{(1)}(k_{f\rho{n}}\rho)\\
R'_F(\rho)_n &= -k_{f\rho{n}}\left[R_{Fn}^+{H_1^{(2)}}(k_{f\rho{n}}\rho)+R_{Fn}^-{H_1^{(1)}}(k_{f\rho{n}}\rho)\right]
\end{align}
The basis vectors reduce to,
\begin{align}
e_{F\rho}(z)_n&=0\\
h_{F\rho}(z)_n&=\frac{-C_{fn}}{\sqrt{Z_{fn}}}Z'_F(z)_{n}=\frac{-C_{fn}}{\sqrt{Z_{fn}}}\frac{n\pi}{z_0}\cos\left(\frac{n\pi}{z_0}z\right)
\end{align}
and where the transverse field components simplify to,
\begin{align}
\vec{e}_{F}(z)_n &= -C_{fn}\sqrt{Z_{fn}}k_{f\rho n}Z_F(z)_n\hat{\phi}= -C_{fn}\sqrt{Z_{fn}}k_{f\rho n}\sin\left(\frac{n\pi}{z_0}z\right)\hat{\phi}\label{eqn:EFbasiscylradialsym}\\
\vec{h}_{F}(z)_n &= \frac{C_{fn}}{\sqrt{Z_{fn}}}k_{f\rho{n}}Z_F(z)_{n}\hat{z}= \frac{C_{fn}}{\sqrt{Z_{fn}}}k_{f\rho{n}}\sin\left(\frac{n\pi}{z_0}z\right)\hat{z}\label{eqn:HFbasiscylradialsym}
\end{align}
where
\begin{align}
C_{fn} = \left[{z_0}k_{f\rho{n}}^2\pi\left[1-\delta_{n,0}\right]\right]^{-\frac{1}{2}}
\end{align}
It is noted that since the electric field has no $\rho$ component, it is $TE_\rho$.  
Since the electric field has only a $\phi$ component, the field is defined as being horizontally polarized.

The total $TE_z$ field components normal to the $\hat{\rho}$ direction, expanded in terms of the basis set reduce to,
\begin{align}
E^F_{\rho}(\rho,z)&=0\\
H^F_{\rho}(\rho,z)&=\sum_{n}{h}^F_{\rho}(z)_{n}\left[R_{Fn}^+{H_{1}^{(2)}}(k_{f\rho{n}}\rho)+R_{F\ell,m,n}^-{H_{1}^{(1)}}(k_{f\rho{n}}\rho)\right]
\end{align}
where the tangential fields expanded in terms of the basis set are given as,
\begin{align}
\vec{E}^F_{\bot}(\rho,z) = \sum_{n}\vec{e}_F(z)_{n}\left[R_{Fn}^+{H_{1}^{(2)}}(k_{f\rho{n}}\rho)+R_{Fn}^-{H_{1}^{(1)}}(k_{f\rho{n}}\rho)\right]\label{eqn:EFperpcyl_radialsym}\\
\vec{H}^F_{\bot}(\rho,z) = \sum_{n}\vec{h}_F(z)_{n}\left[R_{Fn}^+{H_{0}^{(2)}}(k_{f\rho{n}}\rho)+R_{Fn}^-{H_{0}^{(1)}}(k_{f\rho{n}}\rho)\right]\label{eqn:HFperpcyl_radialsym}
\end{align}

